Quantitative Aptitude Practice Questions (With Solutions) Set-27:
List of Practice Questions for Upcoming SSC CGL Tier II Exam was
given here with solutions, candidates those who are preparing for those exams
can use this material
1).If x=√3 - 1/√3 and y=√3 + 1/√3, then the
value of (x/y) + (y2/x) is
a)
√3
b)
3√3
c)
16√3
d)
2√3
2).A number is first decreased by 20%. The
decreased number is then, increased by 20%. The resulting number is less than
the original number by 20. Then the original number is
a)
200
b)
400
c)
500
d)
600
3).If a+b=12, ab=22, then (a2+b2)
is equal to
a)
188
b)
144
c)
34
d)
100
4).Length of a side of a square inscribed in a
circle is a√2 units. The circumference of the circle is
a)
2 πa units
b)
πa units
c)
4 πa units
d)
2a /π units
5).The side of a rhombus are 10 cm are each
and a diagonal measures 16 cm. Area of the rhombus is
a)
96 sq.cm
b)
160 sq.cm
c)
100 sq.cm
d)
40 sq.cm
6).The diameter of the front wheel of an
engine is 2x cm and that of rear wheel is 2y cm. To cover the same distance, find
the number of times the rear wheel will revolve when the front wheel revolves
‘n’ times.
a)
n/xy times
b)
yn/x times
c)
nx/y times
d)
xy/n times
7). The length, breath and height of a cuboid
are in the ratio 3:4:6 and its volume is 576 cm3. The whole surface
of the cuboid is
a)
216 cm2
b)
324 cm2
c)
432cm2
d)
460 cm2
8). The perimeter and length of a rectangle
are 40 m and 12 m respectively. Its breath will be
a)
10m
b)
8m
c)
6m
d)
3m
9). If(x-2) is a factor of x2+3Qx-2Q,
then the value of Q is
a)
2
b)
-2
c)
1
d)
-1
10). If PA and PB are two tangents to a circle
with centre O such that ∠AOB =110
, then ∠APB is

a)
90°
b)
70°
c)
60°
d)
55°
Answers:
1).
b) 2). c) 3).
d) 4). a) 5).
a) 6). c) 7).
c) 8). b) 9).
d) 10). b)
Check
here the Solution for above Aptitude Questions:
1). x=√3-(1/√3)
; y=√3+(1/√3)
x+y =2√3
xy=[√3-(1/√3)] [√3+(1/√3)]
=3-1/3=8/3
x3+y3=(x+y)3-3xy
(x+y)
= (2√3)3-2(8/3)(2√3)
=24√3-√3 =8√3
Now,
(x2/y) + (y2/x)
=x3+y3/xy = 8√3 / (8/3) =3√3
Answer: b)
2). Let
the number be x.
Then 120% of (80% of
x)= x-20
120/100 × 80/100x = x-20
→24x / 25 =x-20
24x=25x-500
∴x =500
Answer: c)
3). a+b=12
ab =22
(a+b)2 =a2+b2+2ab
122 =a2+b2+2×22
144 = a2+b2+44
∴a2+b2=144-44
=100
Answer: d)
4).
DE =0E =a√2 /2 =a√2
Radius =OD =√(OE2+ED2)
= √(a2/2 +
a2/2) = √a2 =a
∴Circumference =2
πr =2 πα
Answer: a)
5).
∆AOB is a right angled
triangle.
Then AB2 =AO2+OB2
100 = AO2+64
AO2 =100 -64
=36
∴AO =6 CM
d1=BD =16
d2=AC=6+6
=12
Area =1/2(d1+d2)
=1/2×16×12 =96cm2
Answer: a)
6). Let the real wheel
rotates N times
Then 2 πr ×n =2xy×N
→N=2πxn / 2πy =xn/y
Answer: c)
7). Let the length,
breath and height be 3x, 4x and 6x respectively
Volume = 3x×4x×6x =576
=72x3 =576
= x3 = 576/72
=x3=8
=x=2
Whole Surface Area = 2×(3x
× 4x + 4x × 6x × 3x)
= 2( 12x2 +
24x2 + 18x2)
= 2 × 54x2
Answer: c)
8). Length l=12m
Let breath be h m
Then Perimeter 2(l+h)
=40
12+h =20-12 =8m
Answer: b)
9). x-2 is a factor of x3+3Qx -2Q
→x=2 is a root of
x2+3Qx-2Q=0
→22+3Q(2)-2Q=0
4+6Q-2Q=0
4 Q = - 4
4Q=-4/4 =-1
Answer: d)
10). ∟APB =180° –∟AOB° =180°-110° =70°
Answer: b)